Printable · GCSE Higher · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Higher
Fifteen questions across the ratio, proportion and rates of change statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Ratio, proportion and rates of change worksheet — GCSE Higher
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- 1.A hiker's distance from the start of the trail, in kilometres, is plotted against time, in minutes, since she set off. At t = 45 minutes she stops to rest, so her distance from the start is neither increasing nor decreasing at that instant. Which statement about the tangent to the distance–time graph at t = 45 minutes is correct?
- 2.There are 84 passengers on a double-decker bus. 35 of the passengers are adults and the rest are children. Write the number of adults as a fraction of the number of children. Give your answer in its simplest form.
- 3.A salt solution has a concentration of 10%. The solution contains 50 g of salt. Work out the total mass of the solution.
- 4.Tap A fills a swimming pool in 6 hours. Tap B pours water twice as fast as tap A. The time taken to fill the pool is inversely proportional to the rate of flow. Work out how long tap B takes to fill the pool.
- 5.y is always the same multiple of x. When x = 6, y = 15. Work out the value of y when x = 10.
- 6.x × y is used to test whether two quantities are in inverse proportion. For the pairs x = 4, y = 15 and x = 6, y = 10, which statement is correct?
- 7.Two quantities are connected by the formula y = 0.8x. Write down the ratio y : x in its simplest form.y = 0.8x
- 8.The number of tickets a group can afford is inversely proportional to the price per ticket. At £4 per ticket, the group can afford 12 tickets. Work out how many tickets the group can afford at £6 per ticket.
- 9.A straight-line graph passes through the origin and the point (4, 10). The line represents y = kx. Write down the ratio x : y in its simplest form.
- 10.A straight-line graph shows the number of pages printed, p, plotted against the time in minutes, t, since a printer started a print job. The line passes through the points (2, 30) and (5, 90). Work out the gradient of the line.
- 11.A laptop priced at £520 is first increased by 15%, and then the new price is decreased by 20%. Work out the final price of the laptop.
- 12.The price of a jacket increases by 50% and then decreases by 50%. Describe the overall change from the original price.
- 13.The ratio of cats to dogs in a shelter is 3:4. There are 18 cats. Work out the total number of animals in the shelter.
- 14.A shade of paint is made by mixing blue paint and white paint. To make 5 litres of the shade, 2 litres of blue paint is used and the rest is white paint. Write the ratio of blue paint to white paint in its simplest form.
- 15.A stall sells bottles of juice. Two bottles cost £3.00, five bottles cost £7.50 and eight bottles cost £12.00. The cost, C pounds, is in a fixed ratio to the number of bottles, n. Write down a formula for C in terms of n.
Answer key
- (c) The tangent is horizontal, so its gradient is 0. — Method: at any point where a distance–time graph is momentarily neither increasing nor decreasing, the tangent to the graph at that point is horizontal, and the gradient of a horizontal line is 0 — this is the instantaneous rate of change at that instant. Working: since the hiker's distance is neither increasing nor decreasing at t = 45 minutes, the tangent there is horizontal, so its gradient is 0. Claiming the tangent is vertical, with an undefined gradient, is the opposite of what the stem says: a vertical tangent would mean the distance was changing infinitely fast at that instant, not that it had stopped changing, and on a distance–time graph it cannot happen at all. Reading the gradient as 45, the time value given in the stem, mistakes a value used to LOCATE the point for the rate of change AT that point. Claiming the gradient cannot be found without also knowing the distance at t = 45 minutes overlooks that 'momentarily stationary' already tells you the rate of change directly, without needing to read any distance value at all. Whenever a stem tells you a quantity is momentarily not changing, that is telling you the instantaneous rate of change directly — it is 0, and no further data is needed to find it.
- (c) 5/7 — First find the number of children: 84 − 35 = 49. The question compares the adults with the children, not with everyone on the bus, so the denominator is 49 and the numerator is 35, giving 35/49. Both parts divide by 7: 35 ÷ 7 = 5 and 49 ÷ 7 = 7. In its simplest form the fraction is 5/7.
- (c) 500 g — Method: a concentration of 10% is the ratio 10:100, and the salt and the solution in the beaker must be in that same ratio, so write 10:100 = 50:m and scale. Working: 50 ÷ 10 = 5, so the salt is 5 times the 10 of the ratio; the solution must be 5 times the 100 of the ratio, giving 5 × 100 = 500 g. Answer: 500 g. The distractors: 5 g comes from working out 10% of 50 g, which treats the 50 g as the whole solution when it is the salt inside it; 450 g comes from scaling correctly and then taking the 50 g of salt away, which gives the mass of water rather than the mass of the whole solution; 5000 g comes from dividing by 0.01 instead of 0.1, that is from writing 10% as 0.01.
- (a) 3 hours — Method: for a fixed pool the rate of flow multiplied by the time taken is constant, so multiplying the rate by a factor divides the time by that same factor. Working: tap B's rate is 2 times tap A's rate, so tap B's time is 6 ÷ 2 = 3 hours. Answer: 3 hours. The distractors: 12 hours comes from multiplying the time by 2 as well, which treats the time as directly proportional to the rate and has the faster tap taking longer; 4 hours comes from reading ‘twice as fast’ additively, as two hours quicker, and working out 6 − 2 instead of scaling the time by a factor of 2; 1.5 hours comes from applying the factor of 2 twice, halving 6 to 3 and then halving again.
- (c) 25 — Find the constant multiplier from the given pair: 15 ÷ 6 = 2.5, so y is always 2.5 times x. When x = 10, y = 10 × 2.5 = 25. 19 comes from assuming an additive relationship instead of a multiplicative one — adding the difference 15 − 6 = 9 onto 10. 4 comes from using the multiplier the wrong way round (6 ÷ 15 = 0.4) and then multiplying by 10. 15 comes from simply repeating the given value of y, without applying the multiplier to the new value of x at all.
- (b) They are in inverse proportion, because x × y = 60 for both pairs. — Testing inverse proportion means checking that x × y is the same for every pair: 4 × 15 = 60 and 6 × 10 = 60, so the quantities are in inverse proportion. Saying they are not in inverse proportion because x + y differs uses addition, which is not the correct test. Saying they are not in inverse proportion because y ÷ x differs uses the test for direct proportion, and finding that it differs tells us nothing about inverse proportion. Saying x × y = 40 for both pairs is an arithmetic slip: 4 × 15 = 60, not 40.
- (c) 4 : 5 — Method: a formula of the form y = kx says that for every 1 of x there are k of y, so the ratio y : x is k : 1, which is then written with whole numbers and cancelled. Working: here k = 0.8, so y : x = 0.8 : 1; multiplying both parts by 10 gives 8 : 10, and dividing both parts by their highest common factor 2 gives 4 : 5. Answer: 4 : 5. The distractors: 5 : 4 comes from writing x : y, reversing the order asked for; 8 : 10 has the decimal cleared but the ratio left unsimplified, and the question asks for the simplest form; 8 : 1 comes from multiplying only the left-hand part by 10 and leaving the 1 unchanged, which changes the relationship instead of keeping it.
- (d) 8 — The product of price and number of tickets is constant: k = 4 × 12 = 48. At £6 per ticket, the number of tickets is 48 ÷ 6 = 8. Getting 18 comes from treating price and tickets as directly proportional and working out 12 × 6 ÷ 4 instead of dividing k by the new price. Getting 12 assumes the number of tickets does not change when the price changes. Getting 6 comes from writing down the new price instead of working out the number of tickets.
- (c) 2 : 5 — The point (4, 10) gives x = 4, y = 10, so x : y = 4 : 10. Dividing both parts by their highest common factor, 2, gives 2 : 5 in simplest form. Inverting the whole ratio gives 5 : 2, which is y : x instead of x : y. Dividing only the x-part by 2 and leaving the y-part as 10 gives 2 : 10, but scaling one part on its own changes the ratio: 2 : 10 is the same as 1 : 5, not 4 : 10. Dividing only the y-part by 2 and leaving the x-part as 4 gives 4 : 5, the same one-sided mistake made on the other part of the ratio.
- (b) 20 — The gradient is the change in p divided by the change in t: (90 − 30) ÷ (5 − 2) = 60 ÷ 3 = 20. Choosing 0.05 comes from dividing the change in t by the change in p, the wrong way round (3 ÷ 60). Choosing −20 comes from subtracting the coordinates in the wrong order for one part of the calculation, for example (30 − 90) ÷ (5 − 2), giving a negative value. Choosing 18 comes from dividing the second p-coordinate by the second t-coordinate directly (90 ÷ 5) instead of using the change between the two points.
- (c) £478.40 — Method: apply the percentage increase, then apply the percentage decrease to the new price. Working: after the increase, the laptop costs £520 × 1.15. Multiplying this result by 0.80 gives the final price, £478.40. Answer: £478.40. £494 comes from combining the two percentages into a single net change (15% − 20% = −5%) and applying it directly, £520 × 0.95 = £494, instead of applying the two changes one after the other. £416 comes from applying only the 20% decrease to the original price, £520 × 0.80 = £416, forgetting the increase entirely. £598 comes from applying only the 15% increase and stopping there, forgetting to apply the decrease at all.
- (c) a decrease of 25% — Method: use multipliers. An increase of 50% is × 1.5 and a decrease of 50% is × 0.5. Working: 1.5 × 0.5 = 0.75, so the final price is 75% of the original. Answer: a decrease of 25%. The distractors: no change comes from assuming +50% and −50% cancel; a decrease of 50% comes from applying only the second change; an increase of 25% has the direction wrong.
- (a) 42 — Find the value of one part: 18 ÷ 3 = 6. Find the number of dogs: 4 × 6 = 24. Add the cats and the dogs to find the total: 18 + 24 = 42. (24 is the number of dogs only, not the total number of animals. 126 comes from multiplying the number of cats by the total number of parts, 18 × 7, instead of finding one part first. 25 comes from adding the ratio numbers 3 and 4 directly to the number of cats.)
- (c) 2:3 — The white paint is 5 − 2 = 3 litres. The ratio of blue paint to white paint is 2 : 3, which has no common factor, so it is already in simplest form. Getting 2 : 5 compares the blue paint to the total amount of shade instead of to the white paint. Getting 3 : 2 has the two parts the wrong way round. Getting 5 : 3 uses the total amount of shade instead of the blue paint as the first part.
- (c) C = 1.5n — Method: a fixed ratio between C and n means C is always the same multiple of n, and that multiple is the cost of one bottle. Working: 3.00 ÷ 2 = 1.5, 7.50 ÷ 5 = 1.5 and 12.00 ÷ 8 = 1.5, so every bottle costs £1.50 and C = 1.5n. Answer: C = 1.5n. The distractors: C = n + 1 comes from subtracting on the first row, 3 − 2 = 1, and adding that difference instead of multiplying; it fits the first row and fails the other two, which is why three rows are given; C = 3n reads the £3.00 as the price of one bottle when it is the price of two; C = n/1.5 divides the number of bottles by the price of one bottle, which works out how many bottles a pound buys instead of what n bottles cost.
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